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Let $X$ be a Banach space. We call a family of bounded operators $(T(t))_{t\in \mathbb{R}}$ a strongly continuous group if it satisfies the properties of the strongly continuous semigroup but for $t\in \mathbb{R}$ instead of $\mathbb{R^+}$. So we can see that every strongly continuous group is a strongly continuous semigroup.

I am just asking if there's a strongly continuous semigroup which cannot be extended to a strongly continuous group.

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The simplest case I know is the translation $T(t)$ semigroup on $L^{2}[0,\infty)$ defined by $(T(t)f)(x)=f(x+t)$. There's no way to invert once you've lost part of the function.

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Consider the Heat semigroup on $L^2(\mathbb{R}^n)$: for $t>0$, $$U(t)g=K(t)\ast g$$ where $$K(t)=(4\pi t)^{-n/2}e^{-|x|/4t}$$ is the heat kernel. $U(t)$ is a strongly continuous semigroupo (of contractions). Let's show that it cannot be extended to a group ($U(t)$, $t\in\mathbb{R}$) on $L^2$. If such a group eventually exists, it should satisfy $$U(t)\circ U(-t)=U(0)=Id$$ for all $t>0$: then, for all $g\in L^2$, $$K(t)\ast U(-t)g=g$$ Applying Fourier transorm we have $e^{-|\xi|^2t}\widehat{U(-t)g}=\hat{g}$, then $$\widehat{U(-t)g}=e^{|\xi|^2t}\hat{g},$$ absurd, since $e^{|\xi|^2t}\hat{g}$ is not, in general, an element of $L^2$.

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  • $\begingroup$ I choose T.A.E 's answer maybe because it is simple, But I also like your example since it concerns maybe the Laplace transform. I would accept both answers if it is possible. $\endgroup$
    – user50618
    Aug 6 '14 at 20:26
  • $\begingroup$ No problem man, i'm happy my answer was useful to you $\endgroup$
    – Capublanca
    Aug 6 '14 at 20:53

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